3.1543 \(\int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=160 \[ \frac{\left (5 a^2 A-2 a b B-A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\sec ^4(c+d x) \left (\left (5 a^2 A-2 a b B+3 A b^2\right ) \sin (c+d x)+2 b (4 a A-b B)\right )}{24 d}+\frac{\left (5 a^2 A-2 a b B-A b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{\sec ^6(c+d x) (a+b \sin (c+d x))^2 (A \sin (c+d x)+B)}{6 d} \]

[Out]

((5*a^2*A - A*b^2 - 2*a*b*B)*ArcTanh[Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^6*(B + A*Sin[c + d*x])*(a + b*Sin[c
+ d*x])^2)/(6*d) + (Sec[c + d*x]^4*(2*b*(4*a*A - b*B) + (5*a^2*A + 3*A*b^2 - 2*a*b*B)*Sin[c + d*x]))/(24*d) +
((5*a^2*A - A*b^2 - 2*a*b*B)*Sec[c + d*x]*Tan[c + d*x])/(16*d)

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Rubi [A]  time = 0.205196, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2837, 821, 778, 199, 206} \[ \frac{\left (5 a^2 A-2 a b B-A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\sec ^4(c+d x) \left (\left (5 a^2 A-2 a b B+3 A b^2\right ) \sin (c+d x)+2 b (4 a A-b B)\right )}{24 d}+\frac{\left (5 a^2 A-2 a b B-A b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{\sec ^6(c+d x) (a+b \sin (c+d x))^2 (A \sin (c+d x)+B)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

((5*a^2*A - A*b^2 - 2*a*b*B)*ArcTanh[Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^6*(B + A*Sin[c + d*x])*(a + b*Sin[c
+ d*x])^2)/(6*d) + (Sec[c + d*x]^4*(2*b*(4*a*A - b*B) + (5*a^2*A + 3*A*b^2 - 2*a*b*B)*Sin[c + d*x]))/(24*d) +
((5*a^2*A - A*b^2 - 2*a*b*B)*Sec[c + d*x]*Tan[c + d*x])/(16*d)

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac{b^7 \operatorname{Subst}\left (\int \frac{(a+x)^2 \left (A+\frac{B x}{b}\right )}{\left (b^2-x^2\right )^4} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}-\frac{b^5 \operatorname{Subst}\left (\int \frac{(a+x) (-5 a A+2 b B-3 A x)}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{6 d}\\ &=\frac{\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac{\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac{\left (b^3 \left (5 a^2 A-A b^2-2 a b B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac{\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac{\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac{\left (5 a^2 A-A b^2-2 a b B\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{\left (b \left (5 a^2 A-A b^2-2 a b B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=\frac{\left (5 a^2 A-A b^2-2 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac{\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac{\left (5 a^2 A-A b^2-2 a b B\right ) \sec (c+d x) \tan (c+d x)}{16 d}\\ \end{align*}

Mathematica [A]  time = 1.48697, size = 242, normalized size = 1.51 \[ \frac{-\frac{3 b \left (-5 a^2 A+2 a b B+A b^2\right ) \left (\left (4 a b^3-6 a^3 b\right ) \tan ^2(c+d x)+\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (\sin (c+d x)+1))-2 \left (a^4-b^4\right ) \tan (c+d x) \sec (c+d x)+2 a^3 b \sec ^2(c+d x)\right )}{16 (a-b) (a+b)}+b \sec ^6(c+d x) (a+b \sin (c+d x))^3 ((b B-a A) \sin (c+d x)-a B+A b)+\frac{1}{4} b \sec ^4(c+d x) (a+b \sin (c+d x))^3 ((2 b B-5 a A) \sin (c+d x)+3 A b)}{6 b d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^6*(a + b*Sin[c + d*x])^3*(A*b - a*B + (-(a*A) + b*B)*Sin[c + d*x]) + (b*Sec[c + d*x]^4*(a + b*
Sin[c + d*x])^3*(3*A*b + (-5*a*A + 2*b*B)*Sin[c + d*x]))/4 - (3*b*(-5*a^2*A + A*b^2 + 2*a*b*B)*((a^2 - b^2)^2*
(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + 2*a^3*b*Sec[c + d*x]^2 - 2*(a^4 - b^4)*Sec[c + d*x]*Tan[c +
d*x] + (-6*a^3*b + 4*a*b^3)*Tan[c + d*x]^2))/(16*(a - b)*(a + b)))/(6*b*(-a^2 + b^2)*d)

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Maple [B]  time = 0.097, size = 396, normalized size = 2.5 \begin{align*}{\frac{{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{B{a}^{2}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{Aab}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{Bab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{Bab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{Bab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Bab\sin \left ( dx+c \right ) }{8\,d}}-{\frac{Bab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{A{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{A{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{A{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{16\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{A{b}^{2}\sin \left ( dx+c \right ) }{16\,d}}-{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{B{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{B{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{12\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/6/d*a^2*A*tan(d*x+c)*sec(d*x+c)^5+5/24/d*a^2*A*tan(d*x+c)*sec(d*x+c)^3+5/16/d*a^2*A*sec(d*x+c)*tan(d*x+c)+5/
16/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*B*a^2/cos(d*x+c)^6+1/3/d*A*a*b/cos(d*x+c)^6+1/3/d*B*a*b*sin(d*x+c)^
3/cos(d*x+c)^6+1/4/d*B*a*b*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*B*a*b*sin(d*x+c)^3/cos(d*x+c)^2+1/8/d*B*a*b*sin(d*x
+c)-1/8/d*B*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*A*b^2*sin(d*x+c)^3/cos(d*x+c)^6+1/8/d*A*b^2*sin(d*x+c)^3/cos(d
*x+c)^4+1/16/d*A*b^2*sin(d*x+c)^3/cos(d*x+c)^2+1/16/d*A*b^2*sin(d*x+c)-1/16/d*A*b^2*ln(sec(d*x+c)+tan(d*x+c))+
1/6/d*B*b^2*sin(d*x+c)^4/cos(d*x+c)^6+1/12/d*B*b^2*sin(d*x+c)^4/cos(d*x+c)^4

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Maxima [A]  time = 1.02698, size = 285, normalized size = 1.78 \begin{align*} \frac{3 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{5} + 12 \, B b^{2} \sin \left (d x + c\right )^{2} - 8 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 8 \, B a^{2} + 16 \, A a b - 4 \, B b^{2} + 3 \,{\left (11 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(sin(d*x + c) + 1) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(sin(d*x + c) - 1
) - 2*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^5 + 12*B*b^2*sin(d*x + c)^2 - 8*(5*A*a^2 - 2*B*a*b - A*b^2)*
sin(d*x + c)^3 + 8*B*a^2 + 16*A*a*b - 4*B*b^2 + 3*(11*A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c))/(sin(d*x + c)^6 -
3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1))/d

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Fricas [A]  time = 1.56532, size = 493, normalized size = 3.08 \begin{align*} \frac{3 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 24 \, B b^{2} \cos \left (d x + c\right )^{2} + 16 \, B a^{2} + 32 \, A a b + 16 \, B b^{2} + 2 \,{\left (3 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, A a^{2} + 16 \, B a b + 8 \, A b^{2} + 2 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d
*x + c)^6*log(-sin(d*x + c) + 1) - 24*B*b^2*cos(d*x + c)^2 + 16*B*a^2 + 32*A*a*b + 16*B*b^2 + 2*(3*(5*A*a^2 -
2*B*a*b - A*b^2)*cos(d*x + c)^4 + 8*A*a^2 + 16*B*a*b + 8*A*b^2 + 2*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^2)
*sin(d*x + c))/(d*cos(d*x + c)^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.4389, size = 309, normalized size = 1.93 \begin{align*} \frac{3 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \,{\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, A a^{2} \sin \left (d x + c\right )^{5} - 6 \, B a b \sin \left (d x + c\right )^{5} - 3 \, A b^{2} \sin \left (d x + c\right )^{5} - 40 \, A a^{2} \sin \left (d x + c\right )^{3} + 16 \, B a b \sin \left (d x + c\right )^{3} + 8 \, A b^{2} \sin \left (d x + c\right )^{3} + 12 \, B b^{2} \sin \left (d x + c\right )^{2} + 33 \, A a^{2} \sin \left (d x + c\right ) + 6 \, B a b \sin \left (d x + c\right ) + 3 \, A b^{2} \sin \left (d x + c\right ) + 8 \, B a^{2} + 16 \, A a b - 4 \, B b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c) + 1)) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x
+ c) - 1)) - 2*(15*A*a^2*sin(d*x + c)^5 - 6*B*a*b*sin(d*x + c)^5 - 3*A*b^2*sin(d*x + c)^5 - 40*A*a^2*sin(d*x
+ c)^3 + 16*B*a*b*sin(d*x + c)^3 + 8*A*b^2*sin(d*x + c)^3 + 12*B*b^2*sin(d*x + c)^2 + 33*A*a^2*sin(d*x + c) +
6*B*a*b*sin(d*x + c) + 3*A*b^2*sin(d*x + c) + 8*B*a^2 + 16*A*a*b - 4*B*b^2)/(sin(d*x + c)^2 - 1)^3)/d